PLEASE SHOW YOUR WORK!!! (solve system of equations by adding, subtracting, or multiplying.)

Mathematics

1 answer:

Aliun [14]3 years ago

5 0

Hey friend, hope I can assist you!
I will solve by elimination <3.
Multiply 3x - 7y = 2 by 2: 6x - 14y = 4
6x - 14y = 4
6x - 9y = 9
5y = 5
Now we have
6x - 14y = 4
5y = 5
Now we want to solve 5y = 5 for y
So simply divide both sides by 5.
5y/5 = 5/5
This gives us one or in other words, y = 1.
Now we want to plug y = 1 into 6x - 14y = 4
So 6x - 14 * 1 = 4
This gives us
6x - 14 = 4
Now add 14 to both sides.
6x - 14 + 14 = 4 + 14
6x = 18
Now divide both sides by 6
6x/6 = 18/6
This gives us 3 so x = 3
Therefore our solutions to this system of equations would be y = 1 and x = 3

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Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$